Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : DEC 2014

By using convolution sum formula i.e.,

$x[n]= (\dfrac {1^n}3)u(n) ∴x[k]= (\dfrac 13)^k u(k). \\ \text {Similarly } \space h[n]= (\dfrac 12)^n u(n) ∴h[n-k]=(\dfrac 12)^{n-k} u(n-k). \\ ∴y[n]= ∑\limits_{-∞}^∞ x[k] × h[n-k]. \\ y[n] = ∑\limits_{k=-∞}^∞.(\dfrac 13)^k \space u(k) (\dfrac 12)^{n-k} u(n-k).$

Since k>0, the lower limit of summation in above …